Factor the following expression: $3$ $x^2$ $-23$ $x+$ $40$
This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(40)} &=& 120 \\ {a} + {b} &=& & & {-23} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $120$ and add them together. The factors that add up to ${-23}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-8}$ and ${b}$ is ${-15}$ $ \begin{eqnarray} {ab} &=& ({-8})({-15}) &=& 120 \\ {a} + {b} &=& {-8} + {-15} &=& -23 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {3}x^2 {-8}x {-15}x +{40} $ Group the terms so that there is a common factor in each group: $ ({3}x^2 {-8}x) + ({-15}x +{40}) $ Factor out the common factors: $ x(3x - 8) - 5(3x - 8) $ Notice how $(3x - 8)$ has become a common factor. Factor this out to find the answer. $(3x - 8)(x - 5)$